Problem-Solving Skills: Colligative Properties

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING CHAPTER 16

Colligative Properties Colligative properties of solutions are properties that depend solely on the number of particles of solute in solution. In other words, these properties depend only on the concentration of solute particles, not on the identity of those particles. Colligative properties result from the interference of solute particles with the motion of solvent molecules. Solute molecules can be either electrolytes or nonelectrolytes. When a nonelectrolyte dissolves, the molecule remains whole in the solution. Glucose and glycerol are examples of nonelectrolyte solutes. Ionic solutes are electrolytes. When they dissolve, they dissociate into multiple particles, or ions. When magnesium chloride dissolves in water, it dissociates as follows. HO

2 MgCl2(s) 9: Mg2⫹(aq) ⫹ 2Cl⫺(aq)

As you can see, when a mole of MgCl2 completely dissociates in solution, it produces 1 mol of Mg2⫹ ions and 2 mol of Cl⫺ ions for a total of 3 mol of solute particles. Because colligative properties depend on the number of particles in solution, 1 mol of MgCl2 in solution should have three times the effect of 1 mol of a nonelectrolyte solute, such as glucose. Two important colligative properties are freezing-point depression and boiling-point elevation. A dissolved solute lowers the freezing point of the solution. The freezing point of a solution differs from that of the pure solvent according to the following equation, in which ⌬tf is the change in freezing point. ⌬tf ⫽ Kf m Kf is a constant that differs for each solvent. Because the freezing point of the solution is lower than that of the solvent alone, Kf is a negative number. The symbol m represents the molality (moles of solute per kilogram of solvent) of the solution. Boiling-point elevation works in the same way. The equation to determine the change in boiling point is as follows. ⌬tb ⫽ Kbm Like Kf , Kb is a constant that differs for each solvent. But unlike Kf , Kb is a positive number because the boiling point of the solution is higher than that of the solvent alone.

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING General Plan for Solving Problems Involving Freezing-Point Depression and Boiling-Point Elevation

1

Mass of solute Convert using the molar mass of the solute.

2

Amount of solute in moles

Determine if solute is an electrolyte or nonelectrolyte. m⫽

3

mol particles kg solvent

4

Amount of particles in solution

Molal concentration of particles in solution, m

Multiply by the molal freezingpoint constant, Kf.

5a

Multiply by the molal boilingpoint constant, Kb.

Freezing-point depression, ⌬tf

5b

Boiling-point elevation, ⌬tb

Add ⌬tf to the normal freezing point of the solvent.

Add ⌬tb to the normal boiling point of the solvent.

6a

6b

Freezing point of solution

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Boiling point of solution

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING Table 16-1 lists freezing-point depression and boiling-point elevation constants for common solvents. TABLE 16-1 Solvent

Normal f.p. Kf

Normal b.p. Kb

Acetic acid

16.6°C

⫺3.90°C/m 117.9°C

3.07°C/m

Camphor

178.8°C

⫺39.7°C/m 207.4°C

5.61°C/m

Ether

⫺116.3°C

⫺1.79°C/m 34.6°C

2.02°C/m

Naphthalene 80.2°C

⫺6.94°C/m 217.7°C

5.80°C/m

Phenol

40.9°C

⫺7.40°C/m 181.8°C

3.60°C/m

Water

0.00°C

⫺1.86°C/m 100.0°C

0.51°C/m

SAMPLE PROBLEM 1 What is the freezing point of a solution of 210.0 g of glycerol, HOCH2CHOHCH2OH, dissolved in 350. g of water? SOLUTION 1. ANALYZE • What is given in the problem? • What are you asked to find?

the formula and mass of solute, and the mass of water used the freezing point of the solution

Items

Data

Identity of solute

glycerol, HOCH2CHOHCH2OH

Particles per mole of solute

1 mol

Identity of solvent

water

Freezing point of solvent

0.00°C

Mass of solvent

350. g

Mass of solute

210.0 g

Molar mass of solute*

92.11 g/mol

Molal concentration of solute particles

?m

Molal freezing-point constant for water

⫺1.86°C/m

Freezing-point depression

?°C


?°C

* determined from the periodic table

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 2. PLAN • What steps are needed to calculate the freezing point of the solution?

Use the molar mass of the solute to determine the amount of solute. Then apply the mass of solvent to calculate the molality of the solution. From the molality, use the molal freezing-point constant for water to calculate the number of degrees the freezing point is lowered. Add this negative value to the normal freezing point.

1 Mass of glycerol in g multiply by the inverse of the molar mass of glycerol

2

Mass of water in g multiply by the conversion factor 1 kg 1000 g

Amount of glycerol in mol the solute is a nonelectrolyte, so the amount of solute equals the amount of particles in solution

3

Mass of water in kg divide the amount of the particles in solution by the mass of the solvent in kilograms


4

Molal concentration of glycerol in water, m multiply by the molal freezing-point constant, Kf , for water

5a Freezing-point depression, ⌬tf add ⌬tf to the normal freezing point of water

6a Freezing point of the glycerol solution given

g H2O ⫻

1 kg ⫽ kg H2O 1000 g

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING

1 molar mass of glycerol given

g glycerol ⫻

calculated above

freezingpoint depression constant

1 mol glycerol 1 ⫺1.86⬚C ⫻ ⫻ ⫽ ⌬tf mol/kg 92.11 g glycerol kg H2O freezing point of H2O

0.00⬚C ⫹

calculated above

⌬tf

⫽ tf

3. COMPUTE 350. g H2O ⫻ 210.0 g glycerol ⫻

1 kg ⫽ 0.350 kg H2O 1000 g

1 mol glycerol 1 ⫻ 92.11 g glycerol 0.350 kg H2O ⫺1.86°C ⫻ ⫽ ⫺12.1°C mol/kg

0.00°C ⫹ (⫺12.1°C) ⫽ ⫺12.1°C 4. EVALUATE • Are the units correct?

Yes; units canceled to give Celsius degrees.

• Is the number of significant figures correct?

Yes; three significant figures is correct because the data had a minimum of three significant figures.

• Is the answer reasonable?

Yes; the calculation can be approximated as 200 ⫼ [90 ⫻ 3(350 ⫼ 1000)] ⫻ ⫺ 2 ⬇ ⫺ 400/30 ⫽ ⫺ 13, which is close to the calculated value.

PRACTICE 1. Determine the freezing point of a solution of 60.0 g of glucose, C6H12O6 , dissolved in 80.0 g of water.

ans: ⫺ 7.74°C

2. What is the freezing point of a solution of 645 g of urea, H2NCONH2 , dissolved in 980. g of water?

ans: ⫺ 20.4°C

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING SAMPLE PROBLEM 2 What is the boiling point of a solution containing 34.3 g of the ionic compound magnesium nitrate dissolved in 0.107 kg of water? SOLUTION 1. ANALYZE • What is given in the problem? • What are you asked to find?

the formula and mass of solute, and the mass of water used the boiling point of the solution

Items

Data

Identity of solute

magnesium nitrate

Equation for the dissociation of the solute

Mg(NO3)2 : Mg2⫹ ⫹ 2NO3⫺

Amount of ions per mole of solute

3 mol

Identity of solvent

water

Boiling point of solvent

100.0°C

Mass of solvent

0.107 kg H2O

Mass of solute

34.3 g

Molar mass of solute

148.32 g/mol


?m

Molal boiling-point constant for solvent

0.51°C/m

Boiling-point depression

?°C

Boiling point of solution

?°C

2. PLAN • What steps are needed to calculate the boiling point of the solution?

Use the molar mass to calculate the amount of solute in moles. Multiply the amount of solute by the number of moles of ions produced per mole of solute. Use the amount of ions with the mass of solvent to compute the molality of particles in solution. Use this effective molality to determine the boiling-point elevation and the boiling point of the solution.

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 1 Mass of Mg(NO3)2 in g multiply by the inverse of the molar mass of Mg(NO3)2

2 Amount of Mg(NO3)2 in mol the solute is an Mass electrolyte, so multiply the amount of solute by the number of particles divide the amount of the per mole of solute particles in solution by the mass of the solvent in 3 kilograms


of water in kg

4

Molal concentration of particles in water, m multiply by the molal boiling-point constant, Kb , for water

5b Boiling-point elevation, ⌬tb add ⌬tb to the normal boiling point of water

6b Boiling point of the Mg(NO3)2 solution given

1 molar mass of Mg(NO3)2

given

1 mol Mg(NO3)2 3 mol particles 1 g Mg(NO3)2 ⫻ ⫻ ⫻ 148.32 g Mg(NO3)2 1 mol Mg(NO3)2 kg H2O molal boiling-point constant for water

⫻ boiling point of H2O

100.0⬚C ⫹

calculated above

⌬tb

0.51⬚C mol/kg

⫽ ⌬tb

⫽ tb

3. COMPUTE 34.3 g Mg(NO3)2 ⫻

1 mol Mg(NO3)2 3 mol particles ⫻ 148.32 g Mg(NO3)2 1 mol Mg(NO3)2 1 0.51°C ⫻ ⫻ ⫽ 3.31°C 0.107 kg H2O mol/kg

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 100.0°C ⫹ 3.31°C ⫽ 103.3°C 4. EVALUATE • Are the units correct? • Is the number of significant figures correct?


Yes; units canceled to give Celsius degrees. Yes; the number of significant figures is correct because the boiling point of water was given to one decimal place. Yes; the calculation can be approximated as [(35 ⫻ 3)/150] ⫻ 5 ⫽ (7/10) ⫻ 5 ⫽ 3.5, which is close to the calculated value for the boilingpoint elevation.

PRACTICE 1. What is the expected boiling point of a brine solution containing 30.00 g of KBr dissolved in 100.00 g of water? 2. What is the expected boiling point of a CaCl2 solution containing 385 g of CaCl2 dissolved in 1.230 ⫻ 103 g of water?

ans: 102.6°C

ans: 104.3°C

SAMPLE PROBLEM 3 A solution of 3.39 g of an unknown compound in 10.00 g of water has a freezing point of ⴚ 7.31°C. The solution does not conduct electricity. What is the molar mass of the compound?

SOLUTION 1. ANALYZE • What is given in the problem?

• What are you asked to find?

the freezing point of the solution, the mass of the dissolved compound, the mass of solvent, and the fact that the solution does not conduct electricity the molar mass of the unknown compound

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING Items

Data

Mass of solute

3.39 g


? g/mol

Identity of solvent

water

Freezing point of solvent

0.00°C

Mass of solvent

10.00 g

Molal freezing-point constant for solvent

⫺ 1.86°C/m

Freezing-point depression

?°C


⫺ 7.31°C


?m

2. PLAN • What steps are needed to calculate the molar mass of the unknown solute?

Determine the molality of the solution from the freezing-point depression. Use the molality and the solute and solvent masses to calculate the solute molar mass.

6a Freezing point of the solution


subtract the normal freezing point of water

divide the mass of the solute by the amount of solute in moles

5a

2

Freezing-point depression, ⌬tf

Amount of solute in mol

multiply by the inverse of the molal freezing-point constant, Kf , for water

the solute is a nonelectrolyte, so the amount of solute equals the amount of particles in solution

4

3

Molal concentration of particles in water, m


multiply the molal concentration by the mass of the water

Mass of water multiply by the conversion in g factor

1 kg 1000 g

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Mass of water in kg

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING

given

freezing point of water

tf ⫺ 0.00⬚C ⫽ ⌬tf

given

g H2O ⫻

calculated above

⌬tf

1 kg ⫽ kg H2O 1000 g

1 molal freezing-point constant for water

⫻

mol/kg ⫺1.86⬚C

calculated above

⫻ kg H2O ⫽ mol solute

given

g solute ⫽ molar mass of solute mol solute calculated above

3. COMPUTE ⫺7.31°C ⫺ 0.00°C ⫽ ⫺7.31°C 1 kg 10.00 g H2O ⫻ ⫽ 0.010 00 kg H2O 1000 g mol/kg ⫺7.31°C ⫻ ⫻ 0.010 00 kg H2O ⫽ 0.039 30 mol solute ⫺1.86°C 3.39 g solute ⫽ 86.3 g/mol 0.039 30 mol solute 4. EVALUATE • Are the units correct? • Is the number of significant figures correct?


Yes; molar mass has units of g/mol. Yes; the number of significant figures is correct because the data had a minimum of three significant figures. Yes; the calculation can be approximated as (4/1) ⫻ (1/100) ⫽ 0.04, which is close to the value of 0.0393 for the amount of solute.

PRACTICE 1. A solution of 0.827 g of an unknown nonelectrolyte compound in 2.500 g of water has a freezing point of ⫺ 10.18°C. Calculate the molar mass of the compound.

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ans: 60.4 g/mol

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 2. A 0.171 g sample of an unknown organic compound is dissolved in ether. The solution has a total mass of 2.470 g. The boiling point of the solution is found to be 36.43°C. What is the molar mass of the organic compound?

ans: 82.1 g/mol

ADDITIONAL PROBLEMS In each of the following problems, assume that the solute is a nonelectrolyte unless otherwise stated. 1. Calculate the freezing point and boiling point of a solution of 383 g of glucose dissolved in 400. g of water. 2. Determine the boiling point of a solution of 72.4 g of glycerol dissolved in 122.5 g of water. 3. What is the boiling point of a solution of 30.20 g of ethylene glycol, HOCH2CH2OH, in 88.40 g of phenol? 4. What mass of ethanol, CH3CH2OH, should be dissolved in 450. g of water to obtain a freezing point of ⫺ 4.5°C? 5. Calculate the molar mass of a nonelectrolyte that lowers the freezing point of 25.00 g of water to ⫺ 3.9°C when 4.27 g of the substance is dissolved in the water. 6. What is the freezing point of a solution of 1.17 g of 1-naphthol, C10H8O, dissolved in 2.00 mL of benzene at 20°C? The density of benzene at 20°C is 0.876 g/mL. Kf for benzene is ⫺ 5.12°C/m, and benzene’s normal freezing point is 5.53°C. 7. The boiling point of a solution containing 10.44 g of an unknown nonelectrolyte in 50.00 g of acetic acid is 159.2°C. What is the molar mass of the solute? 8. A 0.0355 g sample of an unknown molecular compound is dissolved in 1.000 g of liquid camphor at 200.0°C. Upon cooling, the camphor freezes at 157.7°C. Calculate the molar mass of the unknown compound. 9. Determine the boiling point of a solution of 22.5 g of fructose, C6H12O6 , in 294 g of phenol. 10. Ethylene glycol, HOCH2CH2OH, is effective as an antifreeze, but it also raises the boiling temperature of automobile coolant, which helps prevent loss of coolant when the weather is hot. a. What is the freezing point of a 50.0% solution of ethylene glycol in water? b. What is the boiling point of the same 50.0% solution?

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 11. The value of Kf for cyclohexane is ⫺ 20.0°C/m, and its normal freezing point is 6.6°C. A mass of 1.604 g of a waxy solid dissolved in 10.000 g of cyclohexane results in a freezing point of ⫺ 4.4°C. Calculate the molar mass of the solid. 12. What is the expected freezing point of an aqueous solution of 2.62 kg of nitric acid, HNO3 , in a solution with a total mass of 5.91 kg? Assume that the nitric acid is completely ionized. 13. An unknown organic compound is mixed with 0.5190 g of naphthalene crystals to give a mixture having a total mass of 0.5959 g. The mixture is heated until the naphthalene melts and the unknown substance dissolves. Upon cooling, the solution freezes at a temperature of 74.8°C. What is the molar mass of the unknown compound? 14. What is the boiling point of a solution of 8.69 g of the electrolyte sodium acetate, NaCH3COO, dissolved in 15.00 g of water? 15. What is the expected freezing point of a solution of 110.5 g of H2SO4 in 225 g of water? Assume sulfuric acid completely dissociates in water. 16. A compound called pyrene has the empirical formula C8H5 . When 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boiling point of the solution is 85.1°C. Calculate the molar mass of pyrene and determine its molecular formula. The molal boiling-point constant for benzene is 2.53°C/m. Its normal boiling point is 80.1°C. 17. What mass of CaCl2 , when dissolved in 100.00 g of water, gives an expected freezing point of ⫺ 5.0°C; CaCl2 is ionic? What mass of glucose would give the same result? 18. A compound has the empirical formula CH2O. When 0.0866 g is dissolved in 1.000 g of ether, the solution’s boiling point is 36.5°C. Determine the molecular formula of this substance. 19. What is the freezing point of a 28.6% (by mass) aqueous solution of HCl? Assume the HCl is 100% ionized. 20. What mass of ethylene glycol, HOCH2CH2OH, must be dissolved in 4.510 kg of water to result in a freezing point of ⫺ 18.0°C? What is the boiling point of the same solution? 21. A water solution containing 2.00 g of an unknown molecular substance dissolved in 10.00 g of water has a freezing point of ⫺ 4.0°C. a. Calculate the molality of the solution. b. When 2.00 g of the substance is dissolved in acetone instead of in water, the boiling point of the solution is 58.9°C. The normal boiling point of acetone is 56.00°C, and its Kb is 1.71°C/m. Calculate the molality of the solution from this data.

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CHEMFILE MINI-GUIDE TO PROBLEM SOLVING 22. A chemist wants to prepare a solution with a freezing point of ⫺ 22.0°C and has 100.00 g of glycerol on hand. What mass of water should the chemist mix with the glycerol? 23. An unknown carbohydrate compound has the empirical formula CH2O. A solution consisting of 0.515 g of the carbohydrate dissolved in 1.717 g of acetic acid freezes at 8.8°C. What is the molar mass of the carbohydrate? What is its molecular formula? 24. An unknown organic compound has the empirical formula C2H2O. A solution of 3.775 g of the unknown compound dissolved in 12.00 g of water is cooled until it freezes at a temperature of ⫺ 4.72°C. Determine the molar mass and the molecular formula of the compound.

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